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I found this question and answer on a model paper. However I believe the answer is wrong. Can you help me understand please?

Question: 1. Consider the relation R on the set of real numbers R defined by xRy if and only if xy is a rational number. i. Symmetric ii. Reflexive iii. Transitive

Answers given i. As xy = yx for all, x,y ∈ R we have the yx is rational if xy is rational, and so R is symmetric. ii. R is not reflexive as e.g. pi*pi is not rational. iii. R is not transitive as e.g. pi/2 * 2/pi = 1 is rational, while pi/2*pi/2 is irrational.

My concerns: a. I think answer (i) is wrong. Here is why... x = square root of 5 and y = square root of 2, xy = square root of 10 which is irrational. Not for all x,y we have yx rational. Is the model paper answer wrong here?

I think (ii) and (iii) are correct. Right?

Thank you very much!

Sanone
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  • I think answer (i) is wrong $\sqrt{2}\sqrt{5}$ and $\sqrt{5}\sqrt{2}$ are both (equal and) irrational, so neither pair is in the relation $R$. How does that contradict symmetry? – dxiv Apr 04 '17 at 04:15
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    Why did the model paper use pi*pi in the answer (ii) then, because pi is not in the relation R? – Sanone Apr 04 '17 at 04:22
  • $R$ would be reflexive iff $xRx$ for $\forall x \in \mathbb{R},$. But $\pi \in \mathbb{R}$ and $\pi R\pi$ is false, therefore $R$ is not reflexive. – dxiv Apr 04 '17 at 04:25
  • For symmetry, can we use the same argument? Symmetric for all xy? because x and y can be irrational giving xy an irrational. For all xy can't be rational, this is where I can't figure out. Irrational example argument is used to prove "not reflexive" but didn't use the same argument to show "not symmetric". – Sanone Apr 04 '17 at 04:28
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    Reflexivity requires each $x \in \mathbb{R}$ to be in the relation $xRx$ with itself. Symmetry is a conditional statement $xRy \implies yRx,$, which is the same as $xRy \iff yRx,$, but this does not require either side to be true of false, just to be equal for all pairs $x,y \in \mathbb{R},$. – dxiv Apr 04 '17 at 04:32

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Your example does not prove i is wrong. If $x=\sqrt 5, y=\sqrt 2, xy=yx=\sqrt {10}$ then both $xy$ and $yx$ are irrational, so we have $x \not R y$ and $y \not R x$. The iff condition is satisfied because both statements $xRy$ and $yRx$ are false. The point is that because multiplication is commutative we have that either both $xy$ and $yx$ are rational or both are irrational.

Ross Millikan
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    If this is the case, why didn't we use the same argument to pi * pi and say that it is reflexive? The model answer use pi*pi and said not relfexive. (because either xy and yx both rational or both irrational) – Sanone Apr 04 '17 at 04:21
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    Because reflexive requires that $xRx$ for all $x$. $\pi * \pi$ does not prove that $R$ is not symmetric, but it does prove that $R$ is not reflexive. Those are two different questions with different answers. – Ross Millikan Apr 04 '17 at 04:31